Unit 4: Inference for numerical variables Lecture 3: ANOVAtjl13/s101/slides/unit4lec3H.pdf ·...

Unit 4: Inference for numerical variablesLecture 3: ANOVA

Statistics 101

Thomas Leininger

June 10, 2013

Announcements

Announcements

Proposals due tomorrow. Will be returned to you by Wednesday.You MUST complete the proposal process.

A few things to watch out for:

Data is plural, data set is singular.Avoid using population data - if you have population data, youmight consider taking a random sample.Exploratory analysis: should include some summary statistics andsome graphics AND interpretations.If using existing data, find out how your data were collected, anddiscuss the sampling method as well as any possible biases.Scope of inference: generalizability & causality.

Statistics 101 (Thomas Leininger) U4 - L3: ANOVA June 10, 2013 2 / 34

ANOVA Aldrin in the Wolf River

The Wolf River in Tennessee flows past an abandoned site once usedby the pesticide industry for dumping wastes, including chlordane(pesticide), aldrin, and dieldrin (both insecticides).

These highly toxic organic compounds can cause various cancers andbirth defects.

The standard methods to test whether these substances are present ina river is to take samples at six-tenths depth.

But since these compounds are denser than water and their moleculestend to stick to particles of sediment, they are more likely to be found inhigher concentrations near the bottom than near mid-depth.



Data

Aldrin concentration (nanograms per liter) at three levels of depth.

aldrin depth1 3.80 bottom2 4.80 bottom...10 8.80 bottom11 3.20 middepth12 3.80 middepth...20 6.60 middepth21 3.10 surface22 3.60 surface...30 5.20 surface



Exploratory analysis

Aldrin concentration (nanograms per liter) at three levels of depth.bo

ttom

mid

dept

hsu

rfac

e

3 4 5 6 7 8 9

n mean sdbottom 10 6.04 1.58middepth 10 5.05 1.10surface 10 4.20 0.66overall 30 5.1 0 1.37



Research question

Is there a difference between the mean aldrin concentrations amongthe three levels?

To compare means of 2 groups we use a Z or a T statistic.

To compare means of 3+ groups we use a new test calledANOVA and a new statistic called F.



Recap: 2-sample CIs and HTs

n mean sdbottom 10 6.04 1.58middepth 10 5.05 1.10surface 10 4.20 0.66overall 30 5.1 0 1.37

HT: Tdf =(x̄1−x̄2)−null value

SE where SE =

√s2

1n1

+s2

2n2

anddf = min(n1 − 1, n2 − 1)

CI: (x̄1 − x̄2) ± t?df × SE

Application exercise:Perform a HT and construct a CI for each difference.



ANOVA

ANOVA is used to assess whether the mean of the outcome variableis different for different levels of a categorical variable.

H0 : The mean outcome is the same across all categories,

µ1 = µ2 = · · · = µk ,

where µi represents the mean of the outcome for observations incategory i.

HA : At least one mean is different than others.



Conditions

1 The observations should be independent within and betweengroups

If the data are a simple random sample, this condition is satisfied.Carefully consider whether the between-group data isindependent (e.g. no pairing).Always important, but sometimes difficult to check.

2 The observations within each group should be nearly normal.Especially important when the sample sizes are small.

How do we check for normality?

3 The variability across the groups should be about equal.Especially important when the sample sizes differ betweengroups.

How can we check this condition?



z/t test vs. ANOVA - Purpose

z/t test

Compare means from two groupsto see whether they are so farapart that the observed differencecannot reasonably be attributed tosampling variability.

H0 : µ1 = µ2

HA : µ1 , µ2

HA : µ1 < µ2

HA : µ1 > µ2

ANOVA

Compare the means from two ormore groups to see whether theyare so far apart that the observeddifferences cannot all reasonablybe attributed to samplingvariability.

H0 : µ1 = µ2 = · · · = µk

HA : At least one mean is different



z/t test vs. ANOVA - Method

z/t test

Compute a test statistic (a ratio).

z/t =(x̄1 − x̄2) − (µ1 − µ2)

SE(x̄1 − x̄2)

ANOVA

Compute a test statistic (a ratio).

F =variability bet. groupsvariability w/in groups

Large test statistics lead to small p-values.

If the p-value is small enough H0 is rejected, and we concludethat the population means are not equal.



z/t test vs. ANOVA

With only two groups t-test and ANOVA are equivalent, but only ifwe use a pooled standard variance in the denominator of the teststatistic.

With more than two groups, ANOVA compares the samplemeans to an overall grand mean.



Hypotheses

Question

What are the correct hypotheses for testing for a difference betweenthe mean aldrin concentrations among the three levels?

(a) H0 : µB = µM = µS

HA : µB , µM , µS

(b) H0 : µB , µM , µS

HA : µB = µM = µS

(c) H0 : µB = µM = µS

HA : At least one mean is different.

(d) H0 : µB = µM = µS = 0HA : At least one mean is different.

(e) H0 : µB = µM = µS

HA : µB > µM > µS


ANOVA ANOVA and the F test

Test statistic

Does there appear to be a lot of variability within groups? How aboutbetween groups?



ANOVA ANOVA and the F test

F distribution and p-value


In order to be able to reject H0, we need a small p-value, whichrequires a large F statistic.

In order to obtain a large F statistic, variability between samplemeans needs to be greater than variability within sample means.


ANOVA ANOVA output, deconstructed

Df Sum Sq Mean Sq F value Pr(>F)(Group) depth 2 16.96 8.48 6.13 0.0063(Error) Residuals 27 37.33 1.38

Total 29 54.29

Degrees of freedom associated with ANOVA

groups: dfG = k − 1, where k is the number of groups

total: dfT = n − 1, where n is the total sample size

error: dfE = dfT − dfG

dfG = k − 1 = 3 − 1 = 2

dfT = n − 1 = 30 − 1 = 29

dfE = 29 − 2 = 27




Total 29 54.29

Sum of squares between groups, SSG

Measures the variability between groups

SSG =k∑

i=1

ni(x̄i − x̄)2

where ni is each group size, x̄i is the average for each group, x̄ is theoverall (grand) mean.

n meanbottom 10 6.04middepth 10 5.05surface 10 4.2overall 30 5.1

SSG =(10 × (6.04 − 5.1)2

)+

(10 × (5.05 − 5.1)2

)+

(10 × (4.2 − 5.1)2

)= 16.96




Total 29 54.29

Sum of squares total, SSTMeasures the total variability

SST =n∑

i=1

(xi − x̄)

where xi represents each observation in the dataset.

SST = (3.8 − 5.1)2 + (4.8 − 5.1)2 + (4.9 − 5.1)2 + · · ·+ (5.2 − 5.1)2

= (−1.3)2 + (−0.3)2 + (−0.2)2 + · · ·+ (0.1)2

= 1.69 + 0.09 + 0.04 + · · ·+ 0.01

= 54.29




Total 29 54.29

Sum of squares error, SSEMeasures the variability within groups:

SSE = SST − SSG

SSE = 54.29 − 16.96 = 37.33




Total 29 54.29

Mean square error

Mean square error is calculated as sum of squares divided by the de-grees of freedom.

MSG = 16.96/2 = 8.48

MSE = 37.33/27 = 1.38




Total 29 54.29

Test statistic, F valueThe F statistic is the ratio of the between group and within group vari-ability.

F =MSGMSE

F =8.481.38

= 6.14




Total 29 54.29

p-value

p-value is the probability of at least as large a ratio between the “be-tween group” and “within group” variability, if in fact the means of allgroups are equal. It’s calculated as the area under the F curve, withdegrees of freedom dfG and dfE , above the observed F statistic.

0 6.14

dfG = 2 ; dfE = 27



Conclusion

If p-value is small (less than α), reject H0. The data provideconvincing evidence that at least one mean is different from (butwe can’t tell which one).

If p-value is large, fail to reject H0. The data do not provideconvincing evidence that at least one pair of means are differentfrom each other, the observed differences in sample means areattributable to sampling variability (or chance).



Conclusion - in context

Question

What is the conclusion of the hypothesis test for α = 0.05?(p-value = 0.0063)

The data provide convincing evidence that the average aldrinconcentration

(a) is different for all groups.

(b) on the surface is lower than the other levels.

(c) is different for at least one group.

(d) is the same for all groups.


ANOVA Checking conditions

(1) independence

Does this condition appear to be satisfied?



(2) approximately normal


−1.5 −1.0 −0.5 0.0 0.5 1.0 1.5

4

5

6

7

8

9bottom

Theoretical Quantiles

−1.5 −1.0 −0.5 0.0 0.5 1.0 1.5

3.5

4.0

4.5

5.0

5.5

6.0

6.5

middepth


−1.5 −1.0 −0.5 0.0 0.5 1.0 1.5

3.5

4.0

4.5

5.0

surface


3 5 7 9

0

1

2

3

3 5 7

0

1

2

2.5 4.0 5.5

0

1

2

3

4



(3) constant variance


bottomsd=1.58

middepthsd=1.10

surfacesd=0.66

3

4

5

6

7

8

9


Multiple comparisons & Type 1 error rate

Which means differ?

Earlier we concluded that at least one pair of means differ. Thenatural question that follows is “which ones?”

We can do two sample t tests for differences in each possiblepair of groups.

Can you see any pitfalls with this approach?

When we run too many tests, the Type 1 Error rate increases.

This issue is resolved by using a modified significance level.



Multiple comparisons

The scenario of testing many pairs of groups is called multiplecomparisons.

The Bonferroni correction suggests that a more stringentsignificance level is more appropriate for these tests:

α? = α/K

where K is the number of comparisons being considered.

If there are k groups, then usually all possible pairs arecompared and K =

k(k−1)2 .



Determining the modified α

Question

In the aldrin data set depth has 3 levels: bottom, mid-depth, and sur-face. If α = 0.05, what should be the modified significance level for twosample t tests for determining which pairs of groups have significantlydifferent means?

(a) α∗ = 0.05

(b) α∗ = 0.05/2 = 0.025

(c) α∗ = 0.05/3 = 0.0167

(d) α∗ = 0.05/6 = 0.0083



Which means differ?

Question

Based on the box plots below, which means would you expect to besignificantly different?

bottomsd=1.58

middepthsd=1.10

surfacesd=0.66

3

4

5

6

7

8

9 (a) bottom & surface

(b) bottom & mid-depth

(c) mid-depth & surface

(d) bottom & mid-depth;mid-depth & surface

(e) bottom & mid-depth;bottom & surface;mid-depth & surface



Which means differ? (cont.)

If the ANOVA assumption of equal variability across groups issatisfied, we can use the data from all groups to estimate variability:

Estimate any within-group standard deviation with√

MSE, whichis spooled

Use the error degrees of freedom, n − k , for t-distributions

Difference in two means: after ANOVA

SE =

√σ2

1

n1+σ2

2

n2≈

√MSE

n1+

MSEn2



Is there a difference between the average aldrin concentration at thebottom and at mid depth?

n mean sdbottom 10 6.04 1.58middepth 10 5.05 1.10surface 10 4.2 0.66overall 30 5.1 1.37

Df Sum Sq Mean Sq F value Pr(>F)depth 2 16.96 8.48 6.13 0.0063Residuals 27 37.33 1.38Total 29 54.29

TdfE =(x̄bottom − x̄middepth)√

MSEnbottom

+ MSEnmiddepth

T27 =(6.04 − 5.05)√

1.3810 + 1.38

10

=0.990.53

= 1.87

0.05 < p − value < 0.10 (two-sided)

α? = 0.05/3 = 0.0167

Fail to reject H0, the data do not provide convincing evidence of a differencebetween the average aldrin concentrations at bottom and mid depth.



Application exercise:Post-hoc comparison

Is there evidence of a difference between the average aldrin concen-tration at the bottom and at surface?

(a) yes

(b) no


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